若a,b∈R+,求证:1/2(a+b)2+1/4(a+b)≥a√b+b√a

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若a,b∈R+,求证:1/2(a+b)2+1/4(a+b)≥a√b+b√a
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若a,b∈R+,求证:1/2(a+b)2+1/4(a+b)≥a√b+b√a
若a,b∈R+,求证:1/2(a+b)2+1/4(a+b)≥a√b+b√a

若a,b∈R+,求证:1/2(a+b)2+1/4(a+b)≥a√b+b√a
(a+b)^2/2+(a+b)/4 =(a^2+b^2)/2+ab+(a+b)/4 >=2ab+(a+b)/4 =(a+4ab+b+4ab)/4 >=(2根号(4a^2b)+2根号(4ab^2))/4 =(4a根号b+4b根号a)/4 =a根号b+b根号2 所以(a+b)^2/2+(a+b)/4>=a根号b+b根号a